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Question Number 82134 by Raxreedoroid last updated on 18/Feb/20

Commented by Raxreedoroid last updated on 18/Feb/20

$$\mathrm{if}s\mathrm{is}\mathrm{an}\mathrm{arc}\mathrm{for}\mathrm{Circle}\mathrm{C}$$$$\mathrm{and}\mathrm{r}\mathrm{and}\mathrm{Radius}\mathrm{are}\mathrm{Radius}\mathrm{for}\mathrm{C}$$$$\mathrm{then}...$$$$s=\frac{7\pi }{6}$$$$d=\frac{7(\sqrt{6}-\sqrt{2})}{2}$$$$h=3.5$$$$\mathrm{Find}\mathit{r}$$$$r=?$$

Commented by mr W last updated on 18/Feb/20

$$yourquestioniswrongsir!$$$$youshouldknowandseethat$$$$s>d>h$$$$$$$$butyouhave$$$$s=\frac{7\pi }{6}\approx 3.665$$$$d=4$$$$h=\sqrt{6}+\sqrt{2}\approx 3.864$$$$i.e.d>h\stackrel{!}{>}s\Rightarrow thisisimpossible!$$

Commented by Raxreedoroid last updated on 18/Feb/20

$$\mathrm{sorry}!\mathrm{Here}\mathrm{I}\mathrm{edited}\mathrm{the}\mathrm{values}.$$

Commented by mr W last updated on 18/Feb/20

$$todeterminer,onlytwoofs,d,hare$$$$needed.butyouhavegivenallthree$$$$ofthem,thatmeans,oneofthemis$$$$probablywrong.$$

Commented by Raxreedoroid last updated on 19/Feb/20

$$\mathrm{Nothing}\mathrm{wrong}\mathrm{I}\mathrm{can}\mathrm{confirm}$$

Commented by mr W last updated on 19/Feb/20

$$\theta =\frac{s}{r}$$$$d=2r\sin \frac{\theta }{2}=2r\sin \frac{s}{2r}$$$$\Rightarrow \frac{7(\sqrt{6}-\sqrt{2})}{2}=2r\sin \frac{7\pi }{12r}$$$$$$$$h=r\sin \theta =r\sin \frac{s}{r}$$$$\Rightarrow \frac{7}{2}=r\sin \frac{7\pi }{6r}=2r\sin \frac{7\pi }{12r}\cos \frac{7\pi }{12r}$$$$\frac{1}{\sqrt{6}-\sqrt{2}}=\cos \frac{7\pi }{12r}$$$$\frac{\sqrt{6}+\sqrt{2}}{4}=\cos \frac{7\pi }{12r}$$$$\frac{7\pi }{12r}=\frac{\pi }{12}$$$$\Rightarrow r=7$$

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